SATHEE: Application of Derivatives 4 Question 11

Application of Derivatives 4 Question 11

12. Let $f (x) = x^{2} + \frac{1}{x^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1$ . If $h (x) = \frac{f (x)}{g (x)}$ , then the local minimum value of $h (x)$ is

(a) 3

(b) -3

(d) $2 \sqrt{2}$

(2018 Main)

Show Answer

Answer:

Correct Answer: 12. (c)

Solution:

We have,

$\begin{aligned} f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x} \Rightarrow h (x) = \frac{f (x)}{g (x)} \\ ∴ h (x) = \frac{x^{2} + \frac{1}{x^{2}}}{x - \frac{1}{x}} = \frac{x - \frac{1}{x} + 2}{x - \frac{1}{x}} \\ \Rightarrow h (x) = x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} \\ x - \frac{1}{x} > 0, x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} \in [2 \sqrt{2}, \infty) \\ x - \frac{1}{x} < 0, x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} \in (- \infty, 2 \sqrt{2}] \end{aligned}$

$∴$ Local minimum value is $2 \sqrt{2}$ .

पिछला

अगला

गोपनीयता नीति

द्वारा संचालित Prutor@IITK

sathee@iitk.ac.in

SATHEE का मीडिया कवरेज

खेल स्टोर

ऐप स्टोर

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Application of Derivatives 4 Question 11

12. Let f(x)=x2+1x2 and g(x)=x−1x,x∈R−−1,0,1. If h(x)=f(x)g(x), then the local minimum value of h(x) is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

12. Let $f (x) = x^{2} + \frac{1}{x^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1$ . If $h (x) = \frac{f (x)}{g (x)}$ , then the local minimum value of $h (x)$ is